Question

David is driving a steady 26.0 m/sm/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.50 m/s2m/s2 at the instant when David passes. a. How far does Tina drive before passing David?b. What is her speed as she passes him?

Answer 1

a.
`x=540.8m`

b.
`v_f=52(m)/(s)`

Step-by-step explanation:

a. David is moving with constant speed. While Tina is accelerating, we use the equations of uniformly accelerated motion.

For David we have:

`x=vt`

For Tina:

`x=v_0t+(at^2)/(2)\\v_0=0\\x=(at^2)/(2)`

They both travel the same distance from the moment David passes her until she passes David:

`vt=(at^2)/(2)\\t=(2v)/(a)\\t=(2(26(m)/(s)))/(2.5(m)/(s^2))\\t=20.8s`

Knowing this, we can calculate the distance:

`x=vt\\x=26(m)/(s)(20.8s)\\x=540.8m`

b. To calculate her speed when she passes him, we use:

`v_f=v_0+at\\v_f=at\\v_f=(2.5(m)/(s^2))20.8s\\v_f=52(m)/(s)`